Let us analyze each option for the expansion of an ideal gas from the same initial state A:

1. Isothermal expansion (A $\rightarrow$ B): By definition, the temperature remains constant throughout the process. Therefore, $T_A = T_B$.
2. Adiabatic expansion (A $\rightarrow$ C): In an adiabatic expansion, the system does work at the expense of its internal energy ($q = 0$, so $\Delta U = W$). Since the internal energy decreases, the temperature of the ideal gas must drop. This means $T_C < T_A$. Thus, the statement $T_C > T_A$ is incorrect.
3. Entropy change ($\Delta S$): For a reversible adiabatic process, heat exchange is zero ($q_{rev} = 0$), so the entropy change $\Delta S_{\text{adiabatic}} = 0$. For an isothermal expansion, the volume increases, leading to an increase in randomness, so $\Delta S_{\text{isothermal}} > 0$. Therefore, $\Delta S_{\text{isothermal}} > \Delta S_{\text{adiabatic}}$.
4. Work done ($W$): In a P-V diagram, the isothermal curve lies above the adiabatic curve for expansion from the same state. Thus, the area under the isothermal curve is greater, meaning the magnitude of work done by the gas is larger: $|W_{\text{isothermal}}| > |W_{\text{adiabatic}}|$.